Calculus Cheat Sheet Calculus Cheat Sheet
Evaluation Techniques
Limits
Continuous Functions L Hospital s Rule
Definitions
f x f x
If f x is continuous at a then lim f x f a 0
Precise Definition : We say lim f x L if Limit at Infinity : We say lim f x L if we
x a
If lim or lim then,
x a x
x a x a
g x 0 g x
for every 0 there is a 0 such that
can make f x as close to L as we want by
Continuous Functions and Composition
f x f x
whenever 0 x a then f x L .
taking x large enough and positive.
lim lim a is a number, or
f x is continuous at b and lim g x b then
x a x a
g x g x
x a
There is a similar definition for lim f x L Polynomials at Infinity
 Working Definition : We say lim f x L
lim f g x f lim g x f b
x
x a
x ax a
p x and q x are polynomials. To compute
except we require x large and negative.
if we can make f x as close to L as we want
Factor and Cancel
p x
x 2 x 6
by taking x sufficiently close to a (on either side x2 4x 12 lim factor largest power of x in q x out
x
Infinite Limit : We say lim f x if we limlim
q x
of a) without letting x a . x 2 x 2
x a
x2 2xx x 2
of both p x and q x then compute limit.
can make f x arbitrarily large (and positive)
x 6 8
Right hand limit : lim f x L . This has lim4
4
4
by taking x sufficiently close to a (on either side
x 2
x a
x 2 x2 3
3
3x2 43
x2
x2
of a) without letting x a .
the same definition as the limit except it
limlimlim
Rationalize Numerator/Denominator
x
5x 2x2 x 5 x 5 2 2
x2 2
x
requires x a . x
3 x 3 x 3 x
lim lim
There is a similar definition for lim f x
Piecewise Function
x 9 x 9
x a
x2 81 x2 81
3 x
Left hand limit : lim f x L . This has the
x2 5 if x 2
x a except we make f x arbitrarily large and
9 x 1
lim2 g x where g x
limlim
x
same definition as the limit except it requires 1 3x if x 2
x 9 x 9
negative.
x2 81 3 x x 9 3 x
x a .
Compute two one sided limits,
Relationship between the limit and one-sided limits
1 1
lim g x lim x2 5 9
x 2 x 2
lim f x L lim f x lim f x L lim f x lim f x L lim f x L
18 6 108
x a x a
x a x a x a x a
lim g x lim1 3x 7
Combine Rational Expressions x 2 x 2
lim f x lim f x lim f x Does Not Exist
x a
x a x a
One sided limits are different so lim2 g x
x x h
1 1 1 1
x
limlim
h 0 h 0
h x h x h x x h
doesn t exist. If the two one sided limits had
Properties
been equal then lim2 g x would have existed
Assume lim f x and lim g x both exist and c is any number then,
1 h 1 1 x
x a x a
limlim
h 0 h 0
lim f x h x x h x x h x2 and had the same value.
1. lim cf x c lim f x
f x
x a
x ax a
4. lim provided lim g x 0
x a x a
g x lim g x
x a
n
2. lim f x g x lim f x lim g x Some Continuous Functions
n
x ax a x a
5. lim f x lim f x
Partial list of continuous functions and the values of x for which they are continuous.
x ax a
1. Polynomials for all x.
7. cos x and sin x for all x.
n
n
6. lim f x lim f x
3. lim f x g x lim f x lim g x
2. Rational function, except for x s that give
x ax a
x ax a x a
8. tan x and sec x provided
division by zero.
n 33
3. x (n odd) for all x.
Basic Limit Evaluations at
x , , , , ,
n 2 2 2 2
4. x (n even) for all x 0 .
Note : sgn a 1 if a 0 and sgn a 1 if a 0 .
9. cot x and csc x provided
5. ex for all x.
1. lim ex & lim ex 0 5. n even : lim xn
x x x x , 2 , ,0, , 2 ,
6. ln x for x 0 .
2. limln x & limln x 6. n odd : lim xn & lim xn
x x x
x 0
Intermediate Value Theorem
b
7. n even : lim axn bx c sgn a
3. If r 0 then lim 0 Suppose that f x is continuous on [a, b] and let M be any number between f a and f b .
x
x
xr
8. n odd : lim axn bx c sgn a
Then there exists a number c such that a c b and f c M .
4. If r 0 and xr is real for negative x
x
b
9. n odd : lim axn cx d sgn a
then lim 0
x
x
xr
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Chain Rule Variants
Derivatives
The chain rule applied to some specific functions.
Definition and Notation
d nn 1 d
f x h f x
1. f x n f x f x 5. cos f x f x sin f x
If y f x then the derivative is defined to be f x lim .
dx dx
h 0
h
d d
f xf x
2. e f x e 6. tan f x f x sec2 f x
dx dx
If y f x then all of the following are If y f x all of the following are equivalent
d
f x
d
equivalent notations for the derivative. notations for derivative evaluated at x a .
7. sec f (x) f (x) sec f (x) tan f (x)
3. ln f x
dx
dxf x
df dy d
df dy
f x yf x Df x
f a yDf a
f x
d
x a 1
d
dx dx dx
dx dx
8. tan f x
x a x a
4. sin f x f x cos f x 2
dx
1 f x
dx
Interpretation of the Derivative
Higher Order Derivatives
If y f x then, 2. f a is the instantaneous rate of
The Second Derivative is denoted as The nth Derivative is denoted as
1. m f a is the slope of the tangent change of f x at x a .
2 n
d f d f
2 n
f x f x and is defined as f x and is defined as
line to y f x at x a and the 3. If f x is the position of an object at
dx2 dxn
equation of the tangent line at x a is
time x then f a is the velocity of
n n 1
f x f x , i.e. the derivative of the
f x f x , i.e. the derivative of
given by y f a f a x a .
the object at x a .
n 1
first derivative, f x .
the (n-1)st derivative, f x .
Basic Properties and Formulas
Implicit Differentiation
If f x and g x are differentiable functions (the derivative exists), c and n are any real numbers,
Find y if e2 x 9 y x3 y2 sin y 11x . Remember y y x here, so products/quotients of x and y
d
1. c f c f x
5. c 0
will use the product/quotient rule and derivatives of y will use the chain rule. The  trick is to
dx
differentiate as normal and every time you differentiate a y you tack on a y (from the chain rule).
2. f g f x g x d
6. xn nxn 1  Power Rule
After differentiating solve for y .
dx
3. f g f g f g  Product Rule
d
7. f g x f g x g x
e2 x 9 y 2 9y 3x2 y2 2x3 y y cos y y 11
dx
f f g f g
11 2e2 x 9 y 3x2 y2
4.  Quotient Rule
This is the Chain Rule
2e2 x 9 y 9y e2 x 9 y 3x2 y2 2x3 y y cos y y 11 y
g g2
2x3 y 9e2x 9 y cos y
2x3 y 9e2 x 9 y cos y y 11 2e2 x 9 y 3x2 y2
Common Derivatives
d d d
Increasing/Decreasing  Concave Up/Concave Down
x 1 csc x csc x cot x ax ax ln a
dx dx dx
Critical Points
d d d
Concave Up/Concave Down
x c is a critical point of f x provided either
sin x cos x cot x csc2 x ex ex
dx dx dx
1. If f x 0 for all x in an interval I then
1. f c 0 or 2. f c doesn t exist.
d d 1 d 1
cos x sin x sin 1 x ln x , x 0
f x is concave up on the interval I.
dx dx dxx
1 x2
Increasing/Decreasing
2. If f x 0 for all x in an interval I then
d d 1
d 1
tan x sec2 x ln x , x 0
cos 1 x 1. If f x 0 for all x in an interval I then
dx dx x f x is concave down on the interval I.
dx
1 x2
d d 1 f x is increasing on the interval I.
d 1
sec x sec x tan x loga x , x 0
tan 1 x
dx dxx ln a Inflection Points
2. If f x 0 for all x in an interval I then
dx 1 x2
x c is a inflection point of f x if the
f x is decreasing on the interval I.
concavity changes at x c .
3. If f x 0 for all x in an interval I then
f x is constant on the interval I.
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 Calculus Cheat Sheet Calculus Cheat Sheet
Extrema Related Rates
Absolute Extrema Relative (local) Extrema Sketch picture and identify known/unknown quantities. Write down equation relating quantities
1. x c is a relative (or local) maximum of and differentiate with respect to t using implicit differentiation (i.e. add on a derivative every time
1. x c is an absolute maximum of f x
you differentiate a function of t). Plug in known quantities and solve for the unknown quantity.
f x if f c f x for all x near c.
if f c f x for all x in the domain.
Ex. A 15 foot ladder is resting against a wall. Ex. Two people are 50 ft apart when one
2. x c is a relative (or local) minimum of
2. x c is an absolute minimum of f x The bottom is initially 10 ft away and is being starts walking north. The angle changes at
f x if f c f x for all x near c.
1
0.01 rad/min. At what rate is the distance
pushed towards the wall at ft/sec. How fast
4
if f c f x for all x in the domain.
between them changing when 0.5 rad?
is the top moving after 12 sec?
1st Derivative Test
Fermat s Theorem
If x c is a critical point of f x then x c is
If f x has a relative (or local) extrema at
1. a rel. max. of f x if f x 0 to the left
x c , then x c is a critical point of f x .
of x c and f x 0 to the right of x c .
We have 0.01 rad/min. and want to find
2. a rel. min. of f x if f x 0 to the left x is negative because x is decreasing. Using
Extreme Value Theorem
x . We can use various trig fcns but easiest is,
Pythagorean Theorem and differentiating,
If f x is continuous on the closed interval of x c and f x 0 to the right of x c . xx
secsec tan
x2 y2 152 2xx 2y y 0
5050
a,b then there exist numbers c and d so that,
3. not a relative extrema of f x if f x is
1
After 12 sec we have x 10 12 7 and
4 We know 0.05 so plug in and solve.
the same sign on both sides of x c .
1. a c, d b , 2. f c is the abs. max. in
x
so y 152 72 176 . Plug in and solve
sec 0.5 tan 0.5 0.01
a,b , 3. f d is the abs. min. in a,b .
50
2nd Derivative Test for y .
x 0.3112 ft/sec
If x c is a critical point of f x such that
7
1
Finding Absolute Extrema 7 176 y 0 y ft/sec Remember to have calculator in radians!
4
4 176
f c 0 then x c
To find the absolute extrema of the continuous
function f x on the interval a,b use the
1. is a relative maximum of f x if f c 0 .
Optimization
following process.
2. is a relative minimum of f x if f c 0 .
Sketch picture if needed, write down equation to be optimized and constraint. Solve constraint for
1. Find all critical points of f x in a,b .
one of the two variables and plug into first equation. Find critical points of equation in range of
3. may be a relative maximum, relative
variables and verify that they are min/max as needed.
2. Evaluate f x at all points found in Step 1.
minimum, or neither if f c 0 .
Ex. We re enclosing a rectangular field with
Ex. Determine point(s) on y x2 1 that are
3. Evaluate f a and f b .
500 ft of fence material and one side of the
closest to (0,2).
Finding Relative Extrema and/or
4. Identify the abs. max. (largest function
field is a building. Determine dimensions that
Classify Critical Points
value) and the abs. min.(smallest function
will maximize the enclosed area.
1. Find all critical points of f x .
value) from the evaluations in Steps 2 & 3.
2. Use the 1st derivative test or the 2nd
derivative test on each critical point.
22 2
Minimize f d x 0 y 2 and the
Maximize A xy subject to constraint of
constraint is y x2 1. Solve constraint for
Mean Value Theorem
x 2y 500 . Solve constraint for x and plug
If f x is continuous on the closed interval a,b and differentiable on the open interval a,b
x2 and plug into the function.
into area.
2
f b f a x2 y 1 f x2 y 2
A y 500 2y
then there is a number a c b such that f c .
x 500 2y 2
b a
y 1 y 2 y2 3y 3
500 y 2y2
Differentiate and find critical point(s).
Differentiate and find critical point(s).
Newton s Method
3
f 2 y 3 y
A 500 4y y 125
2
f xn
If xn is the nth guess for the root/solution of f x 0 then (n+1)st guess is xn 1 xn
By the 2nd derivative test this is a rel. min. and
By 2nd deriv. test this is a rel. max. and so is
f xn
so all we need to do is find x value(s).
the answer we re after. Finally, find x.
11
provided f xn exists.
x2 3 1 x
x 500 2 125 250
2 2
2
The dimensions are then 250 x 125. 1 3 1 3
The 2 points are then , and , .
2 2
2 2
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 Calculus Cheat Sheet Calculus Cheat Sheet
Standard Integration Techniques
Integrals
Note that at many schools all but the Substitution Rule tend to be taught in a Calculus II class.
Definitions
Definite Integral: Suppose f x is continuous Anti-Derivative : An anti-derivative of f x
bg b
u Substitution : The substitution u g x will convert f g x g x dx f u du using
on a,b . Divide a,b into n subintervals of is a function, F x , such that F x f x .
ag a
*
width x and choose xi from each interval.
Indefinite Integral : f x dx F x c du g x dx . For indefinite integrals drop the limits of integration.
b 2 28
5
where F x is an anti-derivative of f x .
Then f x dx lim f xi* x . Ex. 5x2 cos x3 dx 5x2 cos x3 dx cos u du
3
a n 1 11
i 1
1 8
55
u x3 du 3x2dx x2dx du
3 sin u sin 8 sin 1
33
1
Fundamental Theorem of Calculus
x 1 u 13 1 :: x 2 u 23 8
Variants of Part I :
Part I : If f x is continuous on a,b then
u x
d
x bb
b
f t dt u x f u x
g x f t dt is also continuous on a,b Integration by Parts : udv uv vdu and udv uv vdu . Choose u and dv from
a
dx a
a aa
b
d
x
d
integral and compute du by differentiating u and compute v using v dv .
f t dt v x f v x
and g x f t dt f x .
v x
a dx
dx
5
u x Ex. xe x dx
d
Ex. ln xdx
Part II : f x is continuous on a,b , F x is
f t dt u x f u(x) v x f v(x)
3
v x xx
dx
u x dv e du dx v e 1
an anti-derivative of f x (i.e. F x f x dx ) u ln x dv dx du dx v x
x
x x x x x
xe dx xe e dx xe e c 55 5
5
b
ln xdx x ln x dx x ln x x
then f x dx F b F a .
3
33 3
a
5ln 5 3ln 3 2
Properties
f x g x dx f x dx g x dx cf x dx c f x dx , c is a constant
Products and (some) Quotients of Trig Functions
bbb bb
For sinn x cosm xdx we have the following : For tann x secm xdx we have the following :
f x g x dx f x dx g x dx cf x dx c f x dx , c is a constant
aaa aa
1. n odd. Strip 1 sine out and convert rest to 1. n odd. Strip 1 tangent and 1 secant out and
a bb
convert the rest to secants using
f x dx 0 f x dx f t dt cosines using sin2 x 1 cos2 x , then use
a aa
the substitution u cos x . tan2 x sec2 x 1, then use the substitution
ba
bb
f x dx f x dx 2. m odd. Strip 1 cosine out and convert rest u sec x .
f x dx f x dx
ab
aa
2. m even. Strip 2 secants out and convert rest
to sines using cos2 x 1 sin2 x , then use
ba
If f x g x on a x b then f x dx g x dx the substitution u sin x . to tangents using sec2 x 1 tan2 x , then
ab
3. n and m both odd. Use either 1. or 2. use the substitution u tan x .
b
If f x 0 on a x b then f x dx 0 4. n and m both even. Use double angle 3. n odd and m even. Use either 1. or 2.
a
and/or half angle formulas to reduce the 4. n even and m odd. Each integral will be
b
If m f x M on a x b then m b a f x dx M b a integral into a form that can be integrated. dealt with differently.
a
1 1
Trig Formulas : sin 2x 2sin x cos x , cos2 x 1 cos 2x , sin2 x 1 cos 2x
2 2
Common Integrals
sin5 x
kdx k x c cos udu sin u c tan udu ln sec u c Ex. tan3 x sec5 xdx
Ex. dx
cos3 x
1
(sin2 x)2 sin x
xn dx xn 1 c, n 1 sin udu cos u c secudu ln secu tan u c tan3 xsec5 xdx tan2 xsec4 x tan x sec xdx sin5 x sin4 x sin x
n 1 dxdxdx
cos3 x cos3 x cos3 x
1 u
1 1 1
x 1 dx dx ln x c sec2 udu tan u c du tan c
sec2 x 1 sec4 x tan x sec xdx (1 cos2 x)2 sin x
x
a2 u2 a a
dx u cos x
cos3 x
1 1 1 u
1
dx ln ax b c secu tan udu secu c
du sin c
a
u2 1 u4du u sec x
ax b
(1 u2 )2 1 2u2 u4
a2 u2 a
dudu
u3 u3
cscu cot udu cscu c
1 1
ln udu u ln u u c
sec7 x sec5 x c
7 5
11
sec2 x 2ln cos x cos2 x c
22
csc2 udu cot u c
eu du eu c
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 Calculus Cheat Sheet Calculus Cheat Sheet
Trig Substitutions : If the integral contains the following root use the given substitution and Applications of Integrals
b
formula to convert into an integral involving trig functions.
Net Area : f x dx represents the net area between f x and the
a
a a a
a2 b2x2 x sin b2x2 a2 x sec a2 b2x2 x tan
b b b
x-axis with area above x-axis positive and area below x-axis negative.
cos2 1 sin2 tan2 sec2 1 sec2 1 tan2
16
16 12
2
Ex. dx
cos dd
3
Area Between Curves : The general formulas for the two main cases for each are,
x2 4 9x2 4
sin2
9sin2 2cos
b d
22
x sin dx cos d
33 y f x A upper function lower function dx & x f y A right function left function dy
12csc2 d 12cot c
a c
4 4sin2 4cos2 2 cos
4 9x2
Use Right Triangle Trig to go back to x s. From If the curves intersect then the area of each portion must be found individually. Here are some
3x
sketches of a couple possible situations and formulas for a couple of possible cases.
substitution we have sin so,
Recall x2 x . Because we have an indefinite
2
integral we ll assume positive and drop absolute
value bars. If we had a definite integral we d
need to compute  s and remove absolute value
bars based on that and,
x if x 0
4 9x2
x From this we see that cot . So,
3x
x if x 0
16 4 4 9x2
dxc d
b
cb
x
In this case we have 2cos .
4 9x2
x2 4 9x2
A f y g y dy
A f x g x dx
A f x g x dx g x f x dx
c
a
ac
P x
Partial Fractions : If integrating dx where the degree of P x is smaller than the degree of
Q x
Volumes of Revolution : The two main formulas are V A x dx and V A y dy . Here is
Q x . Factor denominator as completely as possible and find the partial fraction decomposition of
some general information about each method of computing and some examples.
the rational expression. Integrate the partial fraction decomposition (P.F.D.). For each factor in the Rings Cylinders
denominator we get term(s) in the decomposition according to the following table.
A outer radius 2 inner radius 2 A 2 radius width / height
Limits: x/y of right/bot ring to x/y of left/top ring Limits : x/y of inner cyl. to x/y of outer cyl.
Factor in Q x Factor in Q x
Term in P.F.D Term in P.F.D
Horz. Axis use f x , Vert. Axis use f y , Horz. Axis use f y , Vert. Axis use f x ,
A1 A2 Ak
A
k
ax b ax b 2 k g x , A x and dx. g y , A y and dy. g y , A y and dy. g x , A x and dx.
ax b
ax b ax b
ax b
Ax B1 Ak x Bk
Ax B k 1
Ex. Axis : y a 0 Ex. Axis : y a 0 Ex. Axis : y a 0 Ex. Axis : y a 0
k
ax2 bx c
ax2 bx c
ax2 bx c
ax2 bx c
ax2 bx c
A(x2 4) ( Bx C ) (x 1)
7x2 13x Bx C
7x2 13x A
Ex. dx
x 1
( x 1)( x2 4) ( x 1)(x2 4) x2 4 ( x 1)( x2 4)
7 x2 13x 4 3x 16 Set numerators equal and collect like terms.
dxdx
x 1
( x 1)( x2 4) x2 4
7x2 13x A B x2 C B x 4A C
4 3x 16
dx
x 1
Set coefficients equal to get a system and solve
x2 4 x2 4
1 x to get constants.
3
4ln x 1 ln x2 4 8tan
22
radius : a y
A B 7 C B 13 4 A C 0 outer radius : a f x outer radius: a g x radius : a y
Here is partial fraction form and recombined.
width : f y g y
A 4 B 3 C 16
inner radius : a g x inner radius: a f x width : f y g y
An alternate method that sometimes works to find constants. Start with setting numerators equal in
These are only a few cases for horizontal axis of rotation. If axis of rotation is the x-axis use the
previous example : 7x2 13x A x2 4 Bx C x 1 . Chose nice values of x and plug in.
y a 0 case with a 0 . For vertical axis of rotation ( x a 0 and x a 0 ) interchange x and
For example if x 1 we get 20 5A which gives A 4 . This won t always work easily. y to get appropriate formulas.
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 Calculus Cheat Sheet
Average Function Value : The average value
Work : If a force of F x moves an object
b
b
of f x on a x b is favg b 1 f x dx
in a x b , the work done is W F x dx a a
a
Arc Length Surface Area : Note that this is often a Calc II topic. The three basic formulas are,
b b b
L ds SA 2 yds (rotate about x-axis) SA 2 xds (rotate about y-axis)
a a a
where ds is dependent upon the form of the function being worked with as follows.
2 2
2
dy dy
dx
ds 1 dx if y f x , a x b dsdt if x f t , y g t , a t b
dx dt dt
2 2
dx
ds r2 dr d if r f , a b
ds 1 dy if x f y , a y b
d
dy
With surface area you may have to substitute in for the x or y depending on your choice of ds to
match the differential in the ds. With parametric and polar you will always need to substitute.
Improper Integral
An improper integral is an integral with one or more infinite limits and/or discontinuous integrands.
Integral is called convergent if the limit exists and has a finite value and divergent if the limit
doesn t exist or has infinite value. This is typically a Calc II topic.
Infinite Limit
t bb
1. f x dx lim f x dx 2. f x dx lim f x dx
a t a t t
c
3. f x dx f x dx f x dx provided BOTH integrals are convergent.
c
Discontinuous Integrand
bb bt
1. Discont. at a: f x dx lim f x dx 2. Discont. at b : f x dx lim f x dx
a a
t a t t b a
bcb
3. Discontinuity at a c b : f x dx f x dx f x dx provided both are convergent.
aac
Comparison Test for Improper Integrals : If f x g x 0 on a, then,
1. If f x dx conv. then g x dx conv. 2. If g x dx divg. then f x dx divg.
a a a a
1
Useful fact : If a 0 then dx converges if p 1 and diverges for p 1.
p
a
x
Approximating Definite Integrals
b
b a
For given integral f x dx and a n (must be even for Simpson s Rule) define x and
n
a
divide a,b into n subintervals x0, x1 , x1, x2 , & , xn 1, xn with x0 a and xn b then,
b
* ** *
Midpoint Rule : f x dx x f x1 f x2 f xn , xi is midpoint xi 1, xi
a
b
x
Trapezoid Rule : f x dx f x0 2 f x1 2 f x2 2 f xn 1 f xn
a
2
b
x
Simpson s Rule : f x dx f x0 4 f x1 2 f x2 2 f xn 2 4 f xn 1 f xn
a
3
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